Variation Depends on the Interval and Its Endpoint Values
Total variation is a supremum over partitions, so changing the points allowed in a partition can change the answer. On an open interval, partitions do not include the endpoints. On a closed interval, they can include both endpoint values, and those values may differ from the limits approached from inside. Thus the variation on the closed interval can exceed the variation on the open interval, even though the functions agree at every interior point.
Fix finite real numbers \(a<b\). In this tutorial, functions are real-valued. For a function \(f\) defined on an interval \(I\), a finite ordered partition of \(I\) means a finite list \(x_0<\cdots<x_n\) of points of \(I\); its variation sum is \(\sum_{i=1}^n|f(x_i)-f(x_{i-1})|\). The variation \(V_I(f)\) is the supremum of these sums, possibly \(+\infty\). In particular, partitions for \((a,b)\) use only points strictly between \(a\) and \(b\). We assume \(f\) is of bounded variation on \([a,b]\), meaning \(V_{[a,b]}(f)<\infty\). The standard one-sided-limit result for functions of bounded variation gives finite traces \(L=f(a+)=\lim_{x\downarrow a}f(x)\) and \(R=f(b-)=\lim_{x\uparrow b}f(x)\).
The Endpoint Formulas
These identities separate the variation inside the interval from the increments contributed by the endpoint values. They do not say that the endpoint jumps are the only variation near the endpoints: the interior variation already includes all variation captured by partitions using points in \((a,b)\).
Proof. First, we verify a useful description of open-interval variation. Let $$ Q=\sup_{a<y_1<\cdots<y_k<b} \left(|L-f(y_1)|+\sum_{i=2}^{k}|f(y_i)-f(y_{i-1})|+|f(y_k)-R|\right), $$ where the supremum is over all nonempty finite lists of interior points. For any such list, choose interior points \(u_n\downarrow a\) with \(u_n<y_1\) and \(v_n\uparrow b\) with \(v_n>y_k\). The list \(u_n,y_1,\ldots,y_k,v_n\) is an open-interval partition, and its sum tends to the expression defining \(Q\), by the definitions of \(L\) and \(R\). Therefore \(Q\le V_{(a,b)}(f)\). Conversely, for any open-interval partition \(y_1<\cdots<y_k\), its sum is no larger than the expression in parentheses: that expression adds two nonnegative terms. Taking suprema gives \(V_{(a,b)}(f)\le Q\). Hence \(Q=V_{(a,b)}(f)\).
Consider now a partition of \([a,b]\) with at least one interior point, written \(a=x_0<x_1<\cdots<x_n=b\), where \(n\ge2\). The triangle inequality gives $$ \begin{aligned} \sum_{i=1}^n|f(x_i)-f(x_{i-1})| &\le |f(a)-L|+ \left(|L-f(x_1)|+\sum_{i=2}^{n-1}|f(x_i)-f(x_{i-1})|+|f(x_{n-1})-R|\right)\\ &\qquad+|R-f(b)|\\ &\le |f(a)-L|+V_{(a,b)}(f)+|R-f(b)|. \end{aligned} $$ If the partition has no interior points, its sum is \(|f(b)-f(a)|\), which is at most the same bound because \(V_{(a,b)}(f)\ge |L-R|\), as follows by approximating \(L\) and \(R\) with interior values. This proves the upper bound for \(V_{[a,b]}(f)\).
For the reverse bound, fix any list \(y_1<\cdots<y_k\) from the definition of \(Q\). Choose \(u_n\downarrow a\) and \(v_n\uparrow b\) with \(u_n<y_1\) and \(v_n>y_k\). The closed-interval partition \(a,u_n,y_1,\ldots,y_k,v_n,b\) has a sum tending to $$ |f(a)-L|+|L-f(y_1)|+\sum_{i=2}^{k}|f(y_i)-f(y_{i-1})|+|f(y_k)-R|+|R-f(b)|. $$ Taking the supremum over the lists gives the lower bound, since \(Q=V_{(a,b)}(f)\). The closed-interval identity follows.
The same partition argument proves the half-open identities: for \([a,b)\), include \(a\), use interior points up to a point tending to \(b\), and take the supremum; the endpoint contribution at \(a\) is \(|f(a)-L|\), while the contribution at \(b\) is absent. For \((a,b]\), use interior points starting at a point tending to \(a\) and include \(b\); the contribution at \(b\) is \(|R-f(b)|\), while the contribution at \(a\) is absent. In each case the interior sums have supremum \(Q=V_{(a,b)}(f)\), as established above. \(\square\)
Worked Comparisons
Worked Example: Changing Both Endpoint Values
On \([0,1]\), set \(f(x)=x\) for \(0<x<1\), \(f(0)=2\), and \(f(1)=-1\). Compute the variation on all four interval types.
The interior traces are \(L=0\) and \(R=1\). The interior function is increasing, so its variation on \((0,1)\) is \(1\): every partition sum telescopes to its last point minus its first, which is less than \(1\), and points approaching \(0\) and \(1\) make these sums tend to \(1\). Thus $$ V_{(0,1)}(f)=1,\qquad |f(0)-L|=|2-0|=2,\qquad |R-f(1)|=|1-(-1)|=2. $$ The formulas give \(V_{[0,1)}(f)=2+1=3\), \(V_{(0,1]}(f)=1+2=3\), and \(V_{[0,1]}(f)=2+1+2=5\). Check: partitions \(0,u,v,1\), with \(0<u<v<1\), have sum \(2-u+(v-u)+(v+1)=3+2v-2u\), which tends to \(5\) as \(u\downarrow0\), \(v\uparrow1\). No partition can exceed \(5\) by the theorem.
Worked Example: An Interior Jump and Endpoint Changes
On \([0,1]\), define \(f(x)=0\) for \(0<x<\tfrac12\), \(f(x)=3\) for \(\tfrac12\le x<1\), \(f(0)=-2\), and \(f(1)=5\). Compute the open, half-open, and closed variations.
Inside \((0,1)\), the function changes once, from \(0\) to \(3\), so \(V_{(0,1)}(f)=3\). More explicitly, any partition sum is either \(0\) or \(3\), and a partition containing a point below \(\tfrac12\) and a point at or above \(\tfrac12\) has sum \(3\). The endpoint traces are \(L=0\) and \(R=3\). Therefore \(|f(0)-L|=|-2-0|=2\) and \(|R-f(1)|=|3-5|=2\). It follows that $$ V_{[0,1)}(f)=2+3=5,\qquad V_{(0,1]}(f)=3+2=5,\qquad V_{[0,1]}(f)=2+3+2=7. $$ Check: the partition \(0,\tfrac14,\tfrac12,1\) has sum \(|0-(-2)|+|3-0|+|5-3|=2+3+2=7\), so the closed-interval value is attained here.
Worked Example: One Endpoint Value Changes
Let \(g(x)=x\) on \([0,1]\), and let \(h(x)=x\) for \(0<x\le1\) but \(h(0)=4\). Compare their variations.
For both functions, the interior trace at \(0\) is \(0\), the trace at \(1\) is \(1\), and the open-interval variation is \(1\). For \(g\), the endpoint discrepancies are \(|g(0)-0|=0\) and \(|1-g(1)|=0\), giving \(V_{[0,1]}(g)=1\). For \(h\), they are \(|h(0)-0|=|4-0|=4\) and \(|1-h(1)|=|1-1|=0\), giving \(V_{[0,1]}(h)=4+1+0=5\). The half-open variation also changes from \(V_{[0,1)}(g)=1\) to \(V_{[0,1)}(h)=5\), while \(V_{(0,1]}(g)=V_{(0,1]}(h)=1\), because that interval excludes the altered endpoint.
Check: for \(h\), the partition \(0,u,v,1\), where \(0<u<v<1\), has sum \(4-u+(v-u)+(1-v)=5-2u\), which tends to \(5\) as \(u\downarrow0\). For \(g\), every partition sum telescopes to \(1\). These computations agree with the formulas.
A Common Endpoint Error
A frequent mistake is to calculate variation on \([a,b]\) using only the interior trace values, as if the actual endpoint values were forced to equal those traces. In the first worked example, that incorrect calculation gives \(V_{[0,1]}=1\), the variation of \(x\) in the interior. It omits the increments from \(f(0)=2\) to the left trace \(0\), and from the right trace \(1\) to \(f(1)=-1\). The corrected calculation is \(2+1+2=5\). The supremum need not be attained for this correction to matter: partitions can approach the traces from inside, making the endpoint increments arbitrarily close to their limiting sizes.
When using these formulas, first determine the interval on which partitions are allowed. Then compute the interior variation and the one-sided traces, and add an endpoint term only when that endpoint belongs to the interval. If \(a=b\), the interval is a singleton and its variation is \(0\); the formulas above assume \(a<b\).
Check Your Understanding
Use the partition definition and endpoint formulas to justify each answer.
- If \(f(x)=x\) on \((0,2)\), with \(f(0)=3\) and \(f(2)=4\), compute the variations on \((0,2)\), \([0,2)\), \((0,2]\), and \([0,2]\).
- For a bounded-variation function on \([a,b]\), which endpoint discrepancy is omitted from \(V_{[a,b)}(f)\), and why?
- Let \(f(x)=0\) for \(0<x<1\), \(f(1)=2\), and \(f(0)=0\). Compute the open, left-half-open, right-half-open, and closed variations on \([0,1]\).
- Explain why a partition with points strictly inside \((a,b)\) can approximate a contribution involving \(f(a+)\), even though \(a\) is not an allowed partition point for \(V_{(a,b)}(f)\).